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# If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte

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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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15 Dec 2009, 15:48
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of |x – y|?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)
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Re: nonnegative integers - MGMAT Challenge [#permalink]

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15 Dec 2009, 17:04
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axl_oz wrote:
If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of |x – y|?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)

Suppose you got the answer of 2 for the values of $$x$$ and $$y$$ as 4 and 2.

$$2^4+2^2=4^2+2^2$$ --> $$|4-2|=2$$

But if we check for $$y=0$$, we'll get:

$$2^x+2^0=x^2+0^2$$ --> $$2^x+1=x^2$$ --> $$2^x=(x-1)(x+1)$$ --> $$x=3$$

$$2^3+2^0=9=3^2+0^2$$

$$|x-y|=|3-0|=3$$

4 can not be the greatest value as when you increase $$x$$ so as $$x-y$$ to be $$4$$, $$2^x+2^y$$ will always be more than $$x^2+y^2$$.
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Re: nonnegative integers - MGMAT Challenge [#permalink]

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18 Dec 2009, 20:01
good one! i tried backsolving and started with the middle one and then checked one above and one below. got the result.

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Re: nonnegative integers - MGMAT Challenge [#permalink]

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18 Dec 2009, 20:51
it should be 3 , but my way is hit and trial
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Re: nonnegative integers - MGMAT Challenge [#permalink]

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19 Dec 2009, 07:20
Great Problem! Since your trying to find the greatest value of X-Y, you just have to assume that Y=0, like Bunel said and then use the "hit and trial" approach like xcusem... Said. The algebratic approach is great too, but I know for me personally it opens up the opportunity for me to make silly mistakes. So I try to not use it unless necessary.

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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08 Feb 2013, 09:27
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axl_oz wrote:
If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of |x – y|?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)

Since we need to maximize the value of |x – y|, we can do that in two ways...1)make y negative, which is not possible as per the question...2)make y= 0..putting y=0 you will get an equation in x and on hit and trial method u will get the value of x as 3, which will satisfy the equation....
putting x=3 and y=0, we will get the value of |x – y| as 3.

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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20 Mar 2013, 12:16
I have a question for Bunuel: we know that |x−y| = ((x-y)*2)*1/2 = (x*2+y*2-2xy)*1/2

so substituting the value of x*2 +y*2 as 2*x + 2*y in the above equation

I got |x−y| = (2*x + 2*y -2xy)*1/2

then, substituting values for x (taking y=0)...the max value for x can be anything more than 0, coz if you take (x=4) then u'l end up with |x−y| = 4

please can you help me with this problem, where did I go wrong???

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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21 Mar 2013, 03:39
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Ace99 wrote:
I have a question for Bunuel: we know that |x−y| = ((x-y)*2)*1/2 = (x*2+y*2-2xy)*1/2

so substituting the value of x*2 +y*2 as 2*x + 2*y in the above equation

I got |x−y| = (2*x + 2*y -2xy)*1/2

then, substituting values for x (taking y=0)...the max value for x can be anything more than 0, coz if you take (x=4) then u'l end up with |x−y| = 4

please can you help me with this problem, where did I go wrong???

I don't understand the red part above at all...
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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21 Mar 2013, 07:35
sorry...what i meant to say was, if you took the eq: |x−y| = (2*x + 2*y -2xy)*1/2

and substitute the values (x = 4 & y = 0), then we'll end up with |x−y| = (17)*1/2 (which is almost equal to 4)

but before u mentioned the max value of |x−y| = 3.

did i do something wrong??

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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21 Mar 2013, 08:19
Ace99 wrote:
sorry...what i meant to say was, if you took the eq: |x−y| = (2*x + 2*y -2xy)*1/2

and substitute the values (x = 4 & y = 0), then we'll end up with |x−y| = (17)*1/2 (which is almost equal to 4)

but before u mentioned the max value of |x−y| = 3.

did i do something wrong??

First of all, I think you mean 2^x + 2^y -2xy rather than 2*x + 2*y -2xy.

Next, x=4 and y=0 does NOT satisfy 2^x + 2^y = x^2 + y^2, thus these values are not possible.

Hope it's clear.
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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21 Mar 2013, 09:34
Brunel and all,

Is it a rule to apply one value as zero whenever it is given:

1) Both x and y are non-negative integers
2) we need to find the max value of x-y

What if we are asked to find the min ? how do we solve those questions and also, what would be the approach for min and max value of x+y ? Can u guys pls advise?
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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21 Mar 2013, 21:56
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surya167 wrote:
Brunel and all,

Is it a rule to apply one value as zero whenever it is given:

1) Both x and y are non-negative integers
2) we need to find the max value of x-y

What if we are asked to find the min ? how do we solve those questions and also, what would be the approach for min and max value of x+y ? Can u guys pls advise?

Usually, when you are checking for numbers, you do check for 0. It's often a transition point for patterns. Secondly, the question used the term 'non-negative integers' instead of 'positive integers' - this means 0 would probably have a role to play.
There are no such rules but common sense says that we must not ignore 0.

Also, this question doesn't really test max min concepts. It is a direct application of your understanding of exponential relations discussed in the post: http://www.veritasprep.com/blog/2013/01 ... cognition/

Now, when we look at the equation, 2^x + 2^y = x^2 + y^2, some things come to mind:
1. It is not very easy to find values that satisfy this equation.
2. But there must be some values which satisfy since we are looking for a value of |x – y|
3. If x = y = 2, the equation is satisfied since all terms become equal and |x – y| = 0 which is the minimum value of |x – y|.

Usually, the left hand side will be greater than the right hand side (as discussed in the post, 2^n will usually be greater than x^2 except in very few cases). So we must focus on those 'very few cases'. Also, we need to make x and y unequal.

We know (from the post) that 2^4 = 4^2 is one solution so we could put x = 4 while keeping y = 2. The equation will be satisfied and |x – y| = 2

Now, we also know that 2^x < x^2 when x = 3. So is there a solution there as well? The difference between 2^3 and 3^2 is of 1 so can we create a difference of 1 between the other two terms? Sure! If y = 0, then 2^0 = 1 but 0^2 = 0.
So another solution is 2^3 + 2^0 = 3^2 + 0^2.
Here, |x – y| = 3 which is the maximum difference.

The reason we can be sure that there are no other values is that as you go ahead of 4 on the number line, 2^n will be greater than n^2 (again, discussed in the post). So both left hand side terms will be greater than the right hand side terms i.e. 2^x > x^2 and 2^y > y^2. So, for no other values can we satisfy this equation.

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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14 Oct 2013, 12:10
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the expression |x-y| to reach its maximum we need y to be 0. Hence, we need to find X.

therefore, 2^x+2^y=x^2+^2 --> 2^x+1=x^2 what means that x is odd. Only 3 satisfies this equation: 2^3+1=3^2.
Hence, x must be equal 3

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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06 Dec 2013, 10:47
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Bunuel wrote:
misanguyen2010 wrote:
axl_oz wrote:
If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of |x – y|?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)

|x – y| for x and y 2^x + 2^y = x^2 + y^2
A. x= y x=y=1 2 + 2 = 1 + 1 wrong
B. 1 x=2, y=1 4 + 2 = 4 + 1 wrong
C. 2 x=3, y=1 8 + 2 = 9 + 1 right
D. 3 x=4, y =1 16 + 2 = 16 + 1 wrong
E. 4 x=5, y =1 32 + 2 = 25 + 1 wrong

I explained what i confused. Of course I read previous answers and all chose D.
However, from what i found, i chose C. That s why i posted here. I dont know which is wrong in my answer.

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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07 Dec 2013, 05:52
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misanguyen2010 wrote:
Bunuel wrote:
misanguyen2010 wrote:
|x – y| for x and y 2^x + 2^y = x^2 + y^2
A. x= y x=y=1 2 + 2 = 1 + 1 wrong
B. 1 x=2, y=1 4 + 2 = 4 + 1 wrong
C. 2 x=3, y=1 8 + 2 = 9 + 1 right
D. 3 x=4, y =1 16 + 2 = 16 + 1 wrong
E. 4 x=5, y =1 32 + 2 = 25 + 1 wrong

I explained what i confused. Of course I read previous answers and all chose D.
However, from what i found, i chose C. That s why i posted here. I dont know which is wrong in my answer.

To get the greatest value of |x-y| as 3 consider x=3 and y=0. Notice that these values satisfy $$2^x + 2^y = x^2 + y^2$$ --> $$2^3 + 2^0 =9= 3^2 + 0^2$$.

Hope it helps.
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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12 Dec 2013, 02:02
For x>4,y>4
2^x > x^2 ||| 2^y > y^2
Hence maximum we need to check till (x,y) = {0,1,2,3}

By hit and trial .

For x=3,y=1 and x=4,y=2
|x-y| = 2

Draw the graph for 2^x and x^2. The solution becomes simpler .

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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28 Dec 2013, 04:25
Here is how I done it:

1) If |x-y| needs to be max then Y=0, because Y² is only positive
2) Check the answers, those are only integers, you are therefore looking for an integer
3) You have the equation 2^x +1 = X²
4) Use the different choices and you will see that only 3 matches.

Hope it helps!
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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18 Feb 2015, 02:15
If 2^x + 2^y = x^2 + y^2, where x and y are non-negative integers, what is the greatest possible value of |x – y|?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

X=4
Y=1

X-Y = 3

D

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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29 Jul 2015, 21:07
Maximize |x-y| by either making y negative or y = 0. y cannot be negative as given, so make y = 0.

Plug 0 in for y to get x^2 - 1 = 2^x
x^2 -1^2=2^x
If two integers have a median value, then they have a difference of squares.
We already know that they have a difference of squares, so we need to find the median value.
(x+1)(x-1)=2^x
x can only be odd numbers for there to be a median integer.
Plug 3 and you get 4*2 = 2^3. That works, so x=3.

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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01 Sep 2015, 04:33
Bunuel wrote:
axl_oz wrote:
If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of |x – y|?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)

Suppose you got the answer of 2 for the values of $$x$$ and $$y$$ as 4 and 2.

$$2^4+2^2=4^2+2^2$$ --> $$|4-2|=2$$

But if we check for $$y=0$$, we'll get:

$$2^x+2^0=x^2+0^2$$ --> $$2^x+1=x^2$$ --> $$2^x=(x-1)(x+1)$$ --> $$x=3$$

$$2^3+2^0=9=3^2+0^2$$

$$|x-y|=|3-0|=3$$

4 can not be the greatest value as when you increase $$x$$ so as $$x-y$$ to be $$4$$, $$2^x+2^y$$ will always be more than $$x^2+y^2$$.

how to simplify this withlout calculator 2^x = x^2-1

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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte   [#permalink] 01 Sep 2015, 04:33

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